Smith R., Minton R. Calculus

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15-37 SECTION 15.3

Independence of Path and Conservative Vector Fields 1013

43. If F is conservative, then



F · dr = 0 for any closed curve C.

44. If F is conservative, then



F · dr is independent of path.

............................................................

45. Let F(x, y) =

+ y

−y, x. Find a potential function f for

F and carefully note any restrictions on the domain of f. Let C

be the unit circle and show that



F · dr = 2π. Explain why

the Fundamental Theorem for Line Integrals does not apply

to this calculation. Quickly explain how to compute



F · dr

over the circle (x − 2)

+ (y − 3)

= 1.

46. Finish the proof of Theorem 3.3 by showing that if F is

conservative in an open, connected region D ⊂ R

, then



F · dr = 0 for all piecewise-smooth closed curves C lying

in D.

47. Determine whether or not each region is simply-connected.

(a) {(x, y): x

+ y

< 2} (b) {(x, y): 1 < x

+ y

< 2}

48. Determine whether or not each region is simply-connected.

(a) {(x, y): 1 < x < 2} (b) {(x, y): 1 < x

< 2}

APPLICATIONS

49. The Coulomb force for a unit charge at the origin and charge q

at point P

=(x

, y

, z

)isF=

r, where r =



and

r =

x, y, z

. Show that the work done by F to move the

charge q from P

to P

= (x

, y

, z

) is equal to

−

where r



+ y

+ z

and r



+ y

+ z

50. Interpret the result of exercise 49 in the case where (a) P

closer to the origin than P

. (Is the work positive or negative?

Why does this make sense physically?) (b) P

is closer to the

origin than P

and (c) P

and P

are the same distance from

the origin.

51. The work done to increase the temperature of a gas from T

to T

and increase its pressure from P

to P

is given by





dP − RdT



. Here, R is a constant, T is tempera-

ture, P is pressure and C is the path of (P, T ) values as the

changes occur. Compare the work done along the following

two paths. (a) C

consists of the line segment from (P

, T

)

to (P

, T

), followed by the line segment to (P

, T

); (b) C

consists of the line segment from (P

, T

)to(P

, T

), followed

by the line segment to (P

, T

52. Based on your answers in exercise 51, is the force ﬁeld in-

volved in changing the temperature and pressure of the gas

conservative?

53. A vector ﬁeld F satisﬁes F =∇φ (where φ is continuous) at

every point except P, where it is undeﬁned. Suppose that C

is a small closed curve enclosing P, C

is a large closed curve

enclosing C

and C

is a closed curve that does not enclose P.

(See the ﬁgure.) Given that



F · dr = 0, explain why



F ·

dr =



F · dr = 0.

54. The circulation of a ﬂuid with velocity ﬁeld v around the

closed path C is deﬁned by  =



v · dr. For inviscid ﬂow,

 =



v · dv. Show that in this case

 = 0. This is

known as Kelvin’s Circulation Theorem and explains why

small whirlpools in a stream stay coherent and move for peri-

ods of time.

EXPLORATORY EXERCISES

1. For closed curves, we can take advantage of portions of a line

integralthat will equalzero.Forexample,if C is a closedcurve,

explain why you can simplify



(x + y

)dx + (y

+ x)dy



dx + xdy. In general, explain why the f (x)

and g(y) terms can be dropped in the line integral



( f (x) + y

)dx + (x + g(y)) dy. Describe which other

terms can be dropped in the line integral over a closed curve.

Use the example



+ y

+ x

+ cos y)dx

+(y

+ 2xy − x sin y + x

y)dy

to help organize your thinking.

2. In this exercise, we explore a basic principle of physics called

conservation of energy. Start with the work integral



F · dr,

wherethepositionfunctionr(t)isacontinuouslydifferentiable

function of time. Substitute Newton’s second law: F = m

and dr = r



(t) dt = v dt and show that



F · dr = K . Here,

K is kinetic energy deﬁned by K =

mv

and K is the

change of kinetic energy from the initial point of C to the ter-

minal point of C. Next, assume that F is conservative with

F =−∇f , where the function f represents potential energy.

Show that



F · dr =− f where  f equals the change in

potentialenergyfrom the initial point of C to the terminal point

ofC.Concludethatunderthesehypotheses(conservativeforce,

continuous acceleration) the net change in energy K +  f

equals 0. Therefore, K + f is constant.

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1014 CHAPTER 15

Vector Calculus 15-38

15.4 GREEN’S THEOREM

HISTORICAL

NOTES

George Green (1793–1841)

English mathematician who

discovered Green’s Theorem.

Green was self-taught, receiving

only two years of schooling before

going to work in his father’s

bakery at age 9. He continued to

work in and eventually took over

the family mill while teaching

himself mathematics. In 1828, he

published an essay in which he

gave potential functions their

name and applied them to the

study of electricity and

magnetism. This little-read essay

introduced Green’s Theorem and

the so-called Green’s functions

used in the study of partial

differential equations. Green was

admitted to Cambridge University

at age 40 and published several

papers before his early death

from illness. The significance of his

original essay remained unknown

until shortly after his death.

Inthis section, wedevelop a connection between certain line integralsaround a closed curve

in the plane and double integrals over the region enclosed by the curve. At ﬁrst glance, you

might think this a strange and abstract connection, one that only a mathematician could care

about. Actually,the reverseis true; Green’s Theorem is a signiﬁcant result with far-reaching

implications.It is of fundamentalimportanceintheanalysis of ﬂuid ﬂowsandinthetheories

of electricity and magnetism.

Before stating the main result, we brieﬂy deﬁne some terminology. Recall that for a

plane curve C deﬁned parametrically by

C ={(x, y)|x = g(t), y = h(t), a ≤ t ≤ b},

C is closed if its two endpoints are the same, i.e., (g(a), h(a)) = (g(b), h(b)). A curve C is

simple if it does not intersect itself, except at the endpoints. We illustrate a simple closed

curve in Figure 15.28a and a closed curve that is not simple in Figure 15.28b.

FIGURE 15.28a FIGURE 15.28b

Simple closed curve Closed, but not simple curve

We say that a simple closed curve C has positive orientation if the region R enclosed by C

stays to the left of C, as the curve is traversed (as in Figure 15.29a). A curve has negative

orientation if the region R stays to the right of C (as in Figure 15.29b).

FIGURE 15.29a FIGURE 15.29b

Positive orientation Negative orientation

We use the notation



F(x, y) ·dr

to denote a line integral along a simple closed curve C oriented in the positive direction.

We can now state the main result of the section.

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15-39 SECTION 15.4

Green’s Theorem 1015

THEOREM 4.1 (Green’s Theorem)

Let C be a piecewise-smooth, simple closed curve in the plane with positive

orientation and let R be the region enclosed by C, together with C. Suppose that

M(x, y) and N(x, y) are continuous and have continuous ﬁrst partial derivatives in

some open region D, with R ⊂ D. Then,



M(x, y)dx + N(x, y)dy =





∂ N

∂x

−

∂ M

∂y



dA.

You can ﬁnd a general proof of Green’s Theorem in a more advanced text. We prove it

here only for a special case.

a b

: y  g

(x)

: y  g

(x)

FIGURE 15.30a

The region R

PROOF

Here, we assume that the region R can be written in the form

R ={(x, y)|a ≤ x ≤ b and g

(x) ≤ y ≤ g

(x)},

where g

(x) ≤ g

(x), for all x in [a, b], g

(a) = g

(a) and g

(b) = g

(b), as illustrated in

Figure 15.30a. Notice that we can divide C into the two pieces indicated in Figure 15.30a:

C = C

∪C

where C

is the bottom portion of the curve, deﬁned by

={(x, y)|a ≤ x ≤ b, y = g

(x)}

and C

is the top portion of the curve, deﬁned by

={(x, y)|a ≤ x ≤ b, y = g

(x)},

where the orientation is as indicated in the ﬁgure. From the Evaluation Theorem for line

integrals (Theorem 2.4), we then have



M(x, y)dx =



M(x, y)dx +



M(x, y)dx



M(x, g

(x))dx −



M(x, g

(x))dx



[M(x, g

(x)) − M(x, g

(x))]dx, (4.1)

where the minus sign in front of the second integral comes from our traversing C

“back-

ward” (i.e., from right to left). On the other hand, notice that we can write



∂ M

∂y

dA =



(x)

∂ M

∂y

dydx



M(x, y)



y=g

(x)

y=g

(x)

By the Fundamental

Theorem of Calculus



[M(x, g

(x)) − M(x, g

(x))]dx.

Together with (4.1), this gives us



M(x, y)dx =−



∂ M

∂y

dA. (4.2)

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1016 CHAPTER 15

Vector Calculus 15-40

We now assume that we can also write the region R in the form

R ={(x, y)|c ≤ y ≤ d and h

(y) ≤ x ≤ h

(y)},

: x  h

(y)

: x  h

(y)

FIGURE 15.30b

The region R

where h

(y) ≤ h

(y) for all y in [c, d], h

(d) = h

(d). Here, we write

C = C

∪C

, as illustrated in Figure 15.30b. In this case, notice that we can write



N(x, y)dy =



N(x, y)dy +



N(x, y)dy

=−



N(h

(y), y) dy +



N(h

(y), y) dy



[N(h

(y), y) − N(h

(y), y)] dy, (4.3)

where the minus sign in front of the ﬁrst integral accounts for our traversing C

“backward”

(in this case, from top to bottom). Further, notice that



∂ N

∂x

dA =



(y)

∂ N

∂x

dx dy



[N(h

(y), y) − N(h

(y), y)] dy.

Together with (4.3), this gives us



N(x, y)dy =



∂ N

∂x

dA. (4.4)

Adding together (4.2) and (4.4), we have



M(x, y)dx + N(x, y)dy =





∂ N

∂x

−

∂ M

∂y



dA,

as desired.

AlthoughthesigniﬁcanceofGreen’sTheoremliesintheconnectionitprovidesbetween

line integrals and double integrals in more theoretical settings, we illustrate the result in

example 4.1 by using it to simplify the calculation of a line integral.

EXAMPLE 4.1 Using Green’s Theorem

Use Green’s Theorem to rewrite and evaluate



+ y

)dx +3xy

dy, where C

consists of the portion of y = x

from (2, 4) to (0, 0), followed by the line segments

from (0, 0) to (2, 0) and from (2, 0) to (2, 4).

Solution We indicate the curve C and the enclosed region R in Figure 15.31. Notice

that C is a piecewise-smooth, simple closed curve with positive orientation. Further, for

M(x, y) = x

+ y

and N(x, y) = 3xy

, M and N are continuous and have continuous

ﬁrst partial derivatives everywhere. Green’s Theorem then says that



+ y

)dx +3xy

dy =





∂ N

∂x

−

∂ M

∂y





(3y

− 3y

)dA = 0.



FIGURE 15.31

The region R

Notice that in example 4.1, since the integrand of the double integral was zero, eval-

uating the double integral was far easier than evaluating the line integral directly. There is

another simple way of thinking of the line integral in example 4.1. Notice that you can write

this as



F(x, y) ·dr, where F(x, y) =x

+ y

, 3xy

. Notice further that F is conserva-

tive [with potential function f (x, y) =

+ xy

] and so, by Theorem 3.3 in section 15.3,

the line integral of F over any piecewise-smooth, closed curve must be zero.

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15-41 SECTION 15.4

Green’s Theorem 1017

EXAMPLE 4.2 Evaluating a Challenging Line Integral

with Green’s Theorem

Evaluate the line integral



(7y − e

sin x

)dx +[15x − sin(y

+ 8y)]dy, where C is the

circle of radius 3 centered at the point (5, −7), as shown in Figure 15.32.

7

FIGURE 15.32

The region R

Solution First, notice that it will be virtually impossible to evaluate the line integral

directly. (Think about this some, but don’t spend too much time on it!) However, taking

M(x, y) = 7y − e

sin x

and N(x, y) = 15x − sin(y

+ 8y), notice that M and N are

continuous and have continuous ﬁrst partial derivatives everywhere. So, we may apply

Green’s Theorem, which gives us



(7y − e

sin x

)dx +[15x − sin(y

+ 8y)]dy =





∂ N

∂x

−

∂ M

∂y





(15 − 7)dA

= 8



dA = 72π,

where



dA is simply the area inside the region R,



dA = π(3)

= 9π.



If you look at example 4.2 critically, you might suspect that the integrand was chosen

carefullyso thattheline integralwasimpossible to evaluatedirectly,butsothatthe integrand

ofthe double integralwastrivial.That’strue:we did cook up the problem simply to illustrate

the power of Green’s Theorem. Example 4.3 is far less contrived.

EXAMPLE 4.3 Using Green’s Theorem to Evaluate a Line Integral

Evaluate the line integral



+ 6xy)dx + (8x

+ sin y

)dy, where C is the

positively-oriented boundary of the region bounded by the circles of radii 1 and 3,

centered at the origin and lying in the ﬁrst quadrant, as indicated in Figure 15.33.

1 3

FIGURE 15.33

The region R

Solution Notice that since C consists of four distinct pieces, evaluating the line

integral directly by parameterizing the curve is probably not a good choice. On the other

hand, since C is a piecewise-smooth, simple closed curve, we have by Green’s Theorem

that



+ 6xy)dx + (8x

+ sin y

)dy =





∂

∂x

(8x

+ sin y

) −

∂

∂y

+ 6xy)





(16x − 6x)dA =



10xdA,

where R is the region between the two circles and lying in the ﬁrst quadrant. Notice that

this is easy to compute using polar coordinates, as follows:



+ 6xy)dx + (8x

+ sin y

)dy =



10 x



r cos θ



rdrdθ



π/2



(10r cosθ)rdrdθ



π/2

cosθ

10r



r=3

r=1

dθ

− 1

)sin θ



π/2

260



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1018 CHAPTER 15

Vector Calculus 15-42

You should notice that in example 4.3, Green’s Theorem is not a mere convenience;

rather, it is a virtual necessity. Evaluating the line integral directly would prove to be a very

signiﬁcant challenge. (Go ahead and try it to see what we mean.)

Green’s Theorem provides us with a wealth of interesting observations. One of these is

as follows. Suppose that C is a piecewise-smooth, simple closed curve enclosing the region

R. Then, taking M(x, y) = 0 and N(x, y) = x,wehave



xdy=





∂ N

∂x

−

∂ M

∂y



dA =



dA,

which is simply the area of the region R. Alternatively, noticethat if we take M(x, y) =−y

and N(x, y) = 0, we have



−ydx=





∂ N

∂x

−

∂ M

∂y



dA =



dA,

which is again the area of R. Putting these last two results together, we also have



dA =



xdy− ydx. (4.5)

We illustrate this in example 4.4.

FIGURE 15.34

Elliptical region R

EXAMPLE 4.4 Using Green’s Theorem to Find Area

Find the area enclosed by the ellipse

= 1.

Solution First, observe that the ellipse corresponds to the simple closed curve C

deﬁned parametrically by

C ={(x, y)|x = a cos t, y = b sint, 0 ≤ t ≤ 2π},

where a, b > 0. You should also observe that C is smooth and positively oriented. (See

Figure 15.34.) From (4.5), we have that the area A of the ellipse is given by

A =



xdy− ydx =



2π

[(a cost)(b cost) − (b sint)(−a sint)]dt



2π

(abcos

t +ab sin

t) dt = πab.



FIGURE 15.35a

Region with a hole

For simplicity, we often will use the notation ∂R to refer to the boundary of the region

R, oriented in the positive direction. Using this notation, the conclusion of Green’s Theorem

is written as



∂ R

M(x, y)dx + N(x, y)dy =





∂ N

∂x

−

∂ M

∂y



dA.

We can extend Green’s Theorem to the case where a region is not simply-connected (i.e.,

where the region has one or more holes). We must emphasize that when dealing with such

regions, the integration is taken over the entire boundary of the region (not just the outer-

most portion of the boundary!) and that the boundary curve is traversed in the positive

direction, always keeping the region to the left. For instance, for the region R illustrated in

Figure 15.35a with a single hole, notice that the boundary of R,∂R, consists of two separate

curves,C

and C

, where C

is traversed clockwise, in order to keep the orientation positive

on all of the boundary. Since the region is not simply-connected, we may not apply Green’s

Theorem directly. Rather, we ﬁrst make two horizontal slits in the region, as indicated in

Figure 15.35b, dividing R into the two simply-connected regions R

and R

. Notice that

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15-43 SECTION 15.4

Green’s Theorem 1019

we can then apply Green’s Theorem in each of R

and R

separately. Adding the double

integrals over R

and R

gives us the double integral over all of R.Wehave





∂ N

∂x

−

∂ M

∂y



dA =





∂ N

∂x

−

∂ M

∂y



dA +





∂ N

∂x

−

∂ M

∂y





∂ R

M(x, y)dx + N(x, y)dy



∂ R

M(x, y)dx + N(x, y)dy.

Further, since the line integrals over the common portions of ∂ R

and ∂ R

(i.e., the slits)

are traversed in the opposite direction (one way on ∂ R

and the other on ∂ R

), the line

integrals over these portions will cancel out, leaving only the line integrals over C

and C

This gives us





∂ N

∂x

−

∂ M

∂y



dA =



∂ R

M(x, y)dx + N(x, y)dy +



∂ R

M(x, y)dx + N(x, y)dy



M(x, y)dx + N(x, y)dy +



M(x, y)dx + N(x, y)dy



M(x, y)dx + N(x, y)dy.

FIGURE 15.35b

R = R

∪ R

This says that Green’s Theorem also holds for regions with a single hole. Of course, we

can repeat the preceding argument to extend Green’s Theorem to regions with any ﬁnite

number of holes.

FIGURE 15.36

The region R

EXAMPLE 4.5 An Application of Green’s Theorem

For F(x, y) =

+ y

−y, x, show that



F(x, y) ·dr = 2π, for every simple closed

curve C enclosing the origin.

Solution Let C be any simple closed curve enclosing the origin and let C

be the

circle of radius a > 0, centered at the origin (and positively oriented), where a is taken

to be sufﬁciently small so that C

is completely enclosed by C, as illustrated in

Figure 15.36. Further, let R be the region bounded between the curves C and C

(and

including the curves themselves). Applying our extended version of Green’s Theorem

in R,wehave



F(x, y) ·dr −



F(x, y) ·dr



∂ R

F(x, y) ·dr





∂ N

∂x

−

∂ M

∂y







(1)(x

+ y

) − x(2x)

+ y

)

−

(−1)(x

+ y

) + y(2y)

+ y

)





0dA = 0.

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1020 CHAPTER 15

Vector Calculus 15-44

This gives us



F(x, y) ·dr =



F(x, y) ·dr.

Now, we chose C

to be a circle because we can easily parameterize a circle and then

evaluate the line integral around C

explicitly. Notice that C

can be expressed

parametrically by x = a cos t, y = a sin t, for 0 ≤ t ≤ 2π. Noting that on C

+ y

= a

, this leaves us with an integral that we can easily evaluate, as follows:



F(x, y) ·dr =



F(x, y) ·dr =



−y, x·dr



− ydx+ xdy



2π

(−a sint)(−a sint) +(a cos t)(a cost)dt



2π

dt = 2π.



Notice that without Green’s Theorem, proving a result such as that developed in

example 4.5 would be elusive.

Now that we have Green’s Theorem, we are in a position to prove the second half of

Theorem 3.4. For convenience, we restate the theorem here.

THEOREM 4.2

Suppose that M(x, y) and N(x, y) have continuous ﬁrst partial derivatives on a

simply-connected region D. Then,



M(x, y)dx + N(x, y)dy is independent of

path if and only if M

(x, y) = N

(x, y) for all (x, y)inD.

PROOF

Recall that in section 15.3, we proved the ﬁrst part of the theorem: that if



M(x, y)dx + N(x, y)dy is independent of path, then it follows that

(x, y) = N

(x, y) for all (x, y)inD. We now prove that if M

(x, y) = N

(x, y)

for all (x, y)inD, then it follows that the line integral is independent of path. Let S be

any piecewise-smooth closed curve lying in D.IfS is simple and positively oriented, then

since D is simply-connected, the region R enclosed by S is completely contained in D,so

that M

(x, y) = N

(x, y) for all (x, y)inR. From Green’s Theorem, we now have that



M(x, y)dx + N(x, y)dy =





∂ N

∂x

−

∂ M

∂y



dA = 0.

That is, for every piecewise-smooth, simple closed curve S lying in D,wehave



M(x, y)dx + N(x, y)dy = 0. (4.6)

If S is not simple, then it intersects itself one or more times, creating twoor more loops, each

one of which is a simple closed curve. Since the line integral of M(x, y)dx + N(x, y)dy

over each of these is zero by (4. 6), it also follows that



M(x, y)dx + N(x, y)dy = 0.

It now follows from Theorem 3.3 that F(x, y) =M(x, y), N(x, y) must be conservative

in D. Finally, it follows from Theorem 3.1 that



M(x, y)dx + N(x, y)dy is independent

of path.

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15-45 SECTION 15.4

Green’s Theorem 1021

BEYOND FORMULAS

Green’s Theorem is the ﬁrst of three theorems in this chapter that relate different types

of integrals. The alternatives given in these results can be helpful both computationally

and theoretically. Example 4.2 shows how we can evaluate a difﬁcult line integral by

evaluating the equivalent (and simpler) double integral. Perhaps surprisingly, the use

of Green’s Theorem in example 4.5 is probably more important in applications. The

theoretical result can be applied to any relevant problem and results like example 4.5

can sometimes provide important insight into general processes.

EXERCISES 15.4

WRITING EXERCISES

1. Given a line integral to evaluate, brieﬂy describe the circum-

stances under which you should think about using Green’s

Theorem to replace the line integral with a double integral.

Comment on the properties of the curve C and the functions

involved.

2. In example 4.1, Green’s Theorem allowed us to quickly show

thatthe line integralequals 0. Following the example,wenoted

thatthis wasthe line integralfor a conservative force ﬁeld. Dis-

cuss which method (Green’s Theorem, conservative ﬁeld) you

would recommend trying ﬁrst to determine whether a line in-

tegral equals 0.

3. Equation (4.5) shows how to compute area as a line integral.

Using example 4.3 as a guide, explain why we wrote the area



xdy− ydx instead of



xdyor



−ydx.

4. Suppose that you drive a car to a variety of places for shop-

ping and then return home. If your path formed a simple closed

curve, explain how you could use (4.5) to estimate the area en-

closed by your path. (Hint: If x, y represents position, what

does x



, y



 represent?)

In exercises 1–4, evaluate the indicated line integral (a) directly

and (b) using Green’s Theorem.



− y) dx + y

dy, where C is the circle x

+ y

= 1

oriented counterclockwise



+ x)dx + (3x +2xy) dy, where C is the circle

+ y

= 4 oriented counterclockwise



dx − x

dy, where C is the square from (0, 0) to (0, 2)

to (2, 2) to (2, 0) to (0, 0)



− 2x)dx + x

dy, where C is the square from (0, 0) to

(1, 0) to (1, 1) to (0, 1) to (0, 0)

............................................................

In exercises5–20, use Green’s Theorem to evaluatethe indicated

line integral. (Curves are oriented positively unless stated.)



dx − 3x

ydy, where C is the rectangle from (0, 0) to

(3, 0) to (3, 2) to (0, 2) to (0, 0)



dx + x

dy, where C is the rectangle from (−2, 0)

to (3, 0) to (3, 2) to (−2, 2) to (−2, 0)





+ 1

− y



dx + (3x − 4tan y/2)dy, where C is the

portion of y = x

from (−1, 1) to (1, 1), followed by the por-

tion of y = 2 − x

from (1, 1) to (−1, 1)



(xy − e

)dx + (2x

− 4y

)dy, where C is formed by

y = x

and y = 8 − x

oriented clockwise



(tan x − y

)dx + (x

− sin y)dy, where C is the circle

+ y

= 2

10.





√

+ 1 − x



dx + (xy

− y

5/3

)dy, where C is the

circle x

+ y

= 4 oriented clockwise

11.



F ·dr, where F =x

− y, x + y

 and C is formed by

y = x

and y = x

12.



F ·dr, where F =y

+ 3x

y, xy + x

 and C is formed

by y = x

and y = 2x

13.



F ·dr, where F =



− y, e

+ y



and C is formed by

y = 1 − x

and y = 0

14.



F ·dr, where F =x

+ y, x

+ e

 and C is formed by

y = x

and y = 4

15.



dx + 2xdy+ (z − 2) dz, where C is the triangle from

(0, 0, 2) to (2, 2, 2) to (4, 0, 2) to (0, 0, 2)

16.



− ln(x + 1)] dx +





x − y

+ 3x



dy, where C is

formed by x = y

and x = 1

17.



(y sec

x − 2e

)dx + (tan x − 4y

)dy, where C is formed

by x = 4 − y

and x = 0

18.



(zx −2y

)dx + x

zdy+ z

dz, where C is x

+ y

= 4in

the plane z = 2

19.



F ·dr, where F =



−ze

, e

, xe



and C is

+ z

= 1 in the plane y = 0

20.



F ·dr, where F =



−xe

+ y

√

+ z

, 4xy − z



and

C is formed by z = 1 − x

and z = 0 in the plane y = 2

............................................................

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1022 CHAPTER 15

Vector Calculus 15-46

In exercises 21–26, use a line integral to compute the area of the

given region.

21. The ellipse 4x

+ y

= 16 22. The ellipse 4x

+ y

= 4

23. The region bounded by y = x

and y = 4

24. The region bounded by y = x

and y = 2x

25. The region bounded by x

2/3

+ y

2/3

= 1. (Hint: Let x = cos

and y = sin

26. The region bounded by x

2/5

+ y

2/5

= 1

............................................................

27. Use Green’s Theorem to show that the center of mass of the

region bounded by the positive curve C with constant den-

sity is givenby

x =



dy and

y =−



dx,where

A is the area of the region.

28. Use the result of exercise 27 to ﬁnd the center of mass of the

region in exercise 26, assuming constant density.

29. Use the result of exercise 27 to ﬁnd the center of mass of the

region bounded by the curve traced out by t

− t, 1 −t

, for

−1 ≤ t ≤ 1, assuming constant density.

30. Use the result of exercise 27 to ﬁnd the center of mass of the

region bounded by the curve traced out by t

− t, t

− t, for

0 ≤ t ≤ 1, assuming constant density.

31. Use Green’s Theorem to provethe change of variables formula



dA =





∂(x, y)

∂(u,v)



du dv,

where x = x(u,v)and y = y(u,v) are functions with continu-

ous partial derivatives.

32. For F =

+ y

−y, x and C any circle of radius r > 0 not

containing the origin, show that



F · dr = 0.

............................................................

In exercises 33–37, use the technique of example 4.5 to evaluate

the line integral.

33.



F ·dr, where F =



+ y



and C is any posi-

tively oriented simple closed curve enclosing the origin

34.



F ·dr, where F =



− x

+ y

)

−2xy

+ y

)



and C is any

positively oriented simple closed curve enclosing the origin

35.



F ·dr, where F =



+ y



and C is any posi-

tively oriented simple closed curve enclosing the origin

36.



F ·dr, where F =



+ y

−x

+ y



and C is any posi-

tively oriented simple closed curve enclosing the origin

37.



F ·dr, where F =

−y + 1, x

+ (y − 1)

and C is any positively

oriented simple closed curve enclosing (0, 1)

............................................................

38. Where is F(x, y) =



+ y



deﬁned? Show that

= N

everywhere the partial derivatives are deﬁned. If C

is a simple closed curve enclosing the origin, does Green’s

Theorem guarantee that



F ·dr = 0? Explain.

39. For the vector ﬁeld of exercise 38, show that



F ·dr is the

same for all closed curves enclosing the origin.

40. If F(x, y) =



+ y



and C is a simple closed

curve in the fourth quadrant, does Green’s Theorem guarantee

that



F ·dr = 0? Explain.

41. Let C

be the line segment from (0, 1) to (0,0) and C

the triangle from (0, 0) to (1, 1) to (0, 1) to (0, 0). For

F =2xe

, 4x −tan y, compute



F · dr and



F · dr.

Then compute



F · dr, where C

is the bent segment from

(0, 0) to (1, 1) to (0, 1).

42. Compute



(4x − y)dx + (e

− x

y)dy, where C is three

sides of a square from (2, 0) to (2, 2) to (0, 2) to (0, 0).

EXPLORATORY EXERCISE

1. Evaluate



F · dr, where

F =



−y

+ y

)

+ y

)



and C is the circle x

+ y

= a

. Use the result and Green’s

Theorem to show that



−2

+ y

)

dA diverges, where R is

the disk x

+ y

≤ 1.

15.5 CURL AND DIVERGENCE

We haveseen how Green’s Theorem relates the line integral of a function over the boundary

of a plane region R to the double integral of a related function over the region R.In

some cases, the line integral is easier to evaluate, while in other cases, the double integral is

easier. More signiﬁcantly, Green’sTheorem providesuswith a connection between physical

quantities measured on the boundary of a plane region with related quantities in the interior

of the region. The goal of the rest of the chapter is to extend Green’s Theorem to results

that relate triple integrals, double integrals and line integrals. The ﬁrst step is to understand

the vector operations of curl and divergence introduced in this section.